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Two positively charged particles m ₁ and m ₂ have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m ₁=1 amu and m ₂=4 amu] The deBroglie wavelength of m ₁ will be x times of m ₂ . The value of x is _ _ _ _ (nearest integer)

Correct answer

0

Step-by-step solution

For a charged particle accelerated through potential V, the de Broglie wavelength is = h 2mVe . The wavelength is inversely proportional to m . Therefore: ₁ ₂ = m₂ m₁ = 4 1 = 2 . This means ₁ = 2 ₂ , so x = 2 (nearest integer).

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